Control Engineering 1: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

Stability: Routh–Hurwitz & root locus

Stability
all roots of the characteristic polynomial in the left half-plane
Routh array, s¹ entry of a cubic
b1 = (a2a1 − a3a0)/a2
Right half-plane roots
number of sign changes in the first column
Cubic s³ + a₂s² + a₁s + a₀
stable when all a > 0 and a₂·a₁ > a₀
K/(s(s + a)(s + b))
Kmax = ab(a + b), ω = √(ab)
Characteristic equation
1 + K·G(s) = 0
Asymptotes
σa = (Σp − Σz)/(n − m), θ = (2q + 1)·180°/(n − m)
Break-away and break-in points
dK/ds = 0
Magnitude condition
K = 1/|G(s)|
Angle of departure
θd = 180° − Σ∠(from other poles) + Σ∠(from zeros)

Method

  1. Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
  2. Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
  3. Count the sign changes in the first column: each one is a pole in the right half-plane. With K in the array, demand that every first-column entry stays positive.
  4. For a root locus, mark the poles and zeros, the real-axis segments, the asymptotes from the centroid and the break points where dK/ds = 0. Use K = 1/|G(s)| for the gain at a point.

Steady-state error & PID control

System type
number of poles of G(s) at s = 0
Error constants
Kp = lim G(s), Kv = lim s·G(s), Ka = lim s²·G(s) (s → 0)
Step A
ess = A/(1 + Kp)
Ramp A·t
ess = A/Kv
Parabola A·t²/2
ess = A/Ka
PID controller
C(s) = Kp + Ki/s + Kd·s
Second-order target
s² + 2ζωns + ωn²
Overshoot and timing
%OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd, ts ≈ 4/(ζωn)

Method

  1. Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
  2. Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
  3. Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
  4. For a damping or speed specification, write the closed-loop characteristic polynomial and match it with s² + 2ζω_n·s + ω_n².

2Practice problems

1.Laplace transformsMedium
t (s)f(t)t = 0.5f(0.5) = ?

Find f(t) for F(s) = (4s + 13)/(s² + 8s + 16) and evaluate f(0.5).

F(s) = (4s + 13)/(s² + 8s + 16)t = 0.5 s
f(0.5) =
2.Laplace transformsHard
t (s)y(t)t = 0.25y(0.25) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = −8 and y′(0) = −11 using the Laplace transform, and find y(0.25).

y″ + 7y′ + 12y = 0y(0) = −8y′(0) = −11t = 0.25 s
y(0.25) =
3.Block diagramsMedium
R(s)+4K+7s + 5G(s)C(s)2H₁−−

The plant G(s) = 7/(s + 5) has an inner (minor) negative feedback loop with gain H₁ = 2. A controller gain K = 4 and an outer unity negative feedback loop close the system. Find the closed-loop DC gain T(0).

G(s) = 7/(s + 5)H₁ = 2K = 4
T(0) =
4.Block diagramsHard
R(s)+12controller+D(s)+7s + 6plantC(s)−

A proportional controller K = 12 drives the plant G(s) = 7/(s + 6) in a unity negative feedback loop. A disturbance D enters at the plant input (it is added to the controller output). A reference step of size 10 and a disturbance step of size 2 are applied together. Find the steady-state output css.

K = 12G(s) = 7/(s + 6)D = 2R = 10
css =
5.Step responseMedium
t (s)y(t)1ts​ = ?

A system has the transfer function G(s) = 93/(s² + 3s + 93) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 93/(s² + 3s + 93)
ts = s
6.Step responseHard
k = 200 N/mc = 6 N·s/mm = 1 kgF = 25 Nx(t)

A mass m = 1 kg hangs on a spring k = 200 N/m with a viscous damper c = 6 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 25 N is suddenly applied. Find the damping ratio ζ.

m = 1 kgk = 200 N/mc = 6 N·s/mF = 25 N
ζ =
7.Stability: Routh–Hurwitz & root locusMedium
R(s)+K1s(s + 5)(s + 10)C(s)−

The loop in the diagram has the open-loop transfer function L(s) = K/(s(s + 5)(s + 10)). Find the largest gain Kmax for which the closed loop is stable.

L(s) = K/(s(s + 5)(s + 10))
Kmax =
8.Stability: Routh–Hurwitz & root locusHard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K(s + 10)/(s(s + 2)) leaves the real axis between s = 0 and s = −2, circles round and joins the real axis again to the left of the zero. Find the break-in point si.

L(s) = K(s + 10)/(s(s + 2))
si =
9.Steady-state error & PID controlMedium
r(t) = 5t+Ks(s + 10)C(s)−

A unity-feedback loop with G(s) = K/(s(s + 10)) must follow the ramp r(t) = 5t with a steady-state error of 0.1. Find the gain K.

G(s) = K/(s(s + 10))r(t) = 5tess = 0.1
K =
10.Steady-state error & PID controlHard
R(s)+K = 164s(s + 6)C(s)−

A proportional controller K = 16 drives the plant G(s) = 4/(s(s + 6)) in a unity-feedback loop. Find the percent overshoot %OS of the step response.

G(s) = 4/(s(s + 6))K = 16
%OS = %
11.Bode plots & marginsMedium
|H| (dB)0.10.111101010010010001000−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−135−90−450ω = 5

Find the phase ∠G(jω) of G(s) = 6(s + 13)/(s(s + 12)) at ω = 5 rad/s, in degrees.

G(s) = 6(s + 13)/(s(s + 12))ω = 5 rad/s
∠G(jω) = °
12.Bode plots & marginsHard
|H| (dB)0.10.1111010100100−80−60−40−2002040ω (rad/s)φ (°)ω (rad/s)−270−180−900ω180​

A unity-feedback loop has the open-loop transfer function L(s) = 272/(s(s + 4)(s + 8)). Find the gain margin GM in dB.

L(s) = 272/(s(s + 4)(s + 8))
GM = dB

3Solutions

1.Laplace transforms

Answer: f(0.5) = 0.3383

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Factor the denominator

s² + 8s + 16 = (s + 4)²

Partial fractions

4s + 13 = 4(s + 4) + −3
F(s) = 4/(s + 4) − 3/(s + 4)²

Inverse transform

f(t) = 4·e−4t − 3·t·e−4t
f(0.5) = (4 − 3·0.5)·e−4·0.5 = 0.3383

Answer

f(0.5) = 0.3383
2.Laplace transforms

Answer: y(0.25) = −7.436

Laplace transform: F(s) = ∫₀∞ f(t)·e−st dt turns derivatives into multiplications by s: ℒ{f′} = sF(s) − f(0), ℒ{f″} = s²F(s) − s·f(0) − f′(0). Standard pairs: 1 ↔ 1/s, e−at ↔ 1/(s + a), t·e−at ↔ 1/(s + a)², sin ωt ↔ ω/(s² + ω²), cos ωt ↔ s/(s² + ω²), and a shift s → s + σ multiplies the time function by e−σt. To invert a rational F(s), split it into partial fractions: the residue of a simple pole p is A = [(s − p)·F(s)] at s = p. Final value theorem: f(∞) = lims→0 s·F(s), valid only when all poles of s·F(s) lie in the left half-plane. Initial value theorem: f(0⁺) = lims→∞ s·F(s).

Transform the equation

s²Y − (−8)s − (−11) + 7(sY − (−8)) + 12Y = 0
Y(s) = (−8s − 67)/((s + 3)(s + 4))

Partial fractions

A = −43/1 = −43
B = −35/(−1) = 35

Result

y(t) = −43·e−3t + 35·e−4t
y(0.25) = −7.436

Answer

y(0.25) = −7.436

Check: y(0) = A + B = −8.

3.Block diagrams

Answer: T(0) = 0.5957

Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).

Inner loop

Gᵢ = G/(1 + G·H₁) = 7/(s + 5 + 7·2) = 7/(s + 19)

Outer loop

T(s) = K·Gᵢ/(1 + K·Gᵢ) = 28/(s + 19 + 28) = 28/(s + 47)

Result

T(0) = 28/47 = 0.5957

Answer

T(0) = 0.5957
4.Block diagrams

Answer: css = 9.489

Block-diagram algebra: blocks in series multiply (G₁·G₂), blocks in parallel add (G₁ + G₂), and a loop with forward path G and feedback path H reduces to T = G/(1 + G·H) for negative feedback and G/(1 − G·H) for positive feedback. Reduce inner loops first, then work outwards. The closed-loop poles are the roots of the characteristic equation 1 + G(s)H(s) = 0. For a stable loop, a step input of size A gives the steady-state output A·T(0); with unity feedback the steady-state error to a step is A/(1 + Kp) with Kp = G(0).

Transfer functions

C/D = G/(1 + K·G) = 7/(s + 6 + 84)
C/R = K·G/(1 + K·G) = 84/(s + 6 + 84)

Steady state

css = 10·84/90 + 2·7/90 = 9.489

Answer

css = 9.489

A larger K reduces the effect of the disturbance, but never to zero with a proportional controller.

5.Step response

Answer: ts = 2.667 s

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 93 → ωₙ = 9.644 rad/s
2ζωₙ = 3 → ζ = 3/(2·9.644) = 0.1555

Result

ζωₙ = 1.5 → ts ≈ 4/1.5 = 2.667 s

Answer

ts = 2.667 s
6.Step response

Answer: ζ = 0.2121

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(200/1) = 14.14 rad/s
ζ = c/(2√(km)) = 6/(2·√(200·1)) = 0.2121

Result

ζ = 0.2121

Answer

ζ = 0.2121
7.Stability: Routh–Hurwitz & root locus

Answer: Kmax = 750

Routh–Hurwitz criterion: a linear system is stable when every root of its characteristic polynomial lies in the left half of the s-plane. Write the coefficients alternately into the first two rows of the Routh array. Each entry of a new row comes from the two rows above it; for a cubic a₃s³ + a₂s² + a₁s + a₀ the s¹ entry is b₁ = (a₂·a₁ − a₃·a₀)/a₂. The number of sign changes in the first column equals the number of roots in the right half-plane. With a gain K in the coefficients, every first-column entry must stay positive: the gain that makes the s¹ entry zero is the stability limit, and the s² row then gives the frequency of the sustained oscillation.

Characteristic equation

s³ + 15s² + 50s + K = 0

Routh array

s³: 1, 50
s²: 15, K
s¹: (15·50 − K)/15
s⁰: K

Conditions

Every entry of the first column must be positive.

K > 0
15·50 − K > 0 → K < 750

Answer

Kmax = 750

At K = 750 the closed loop oscillates at ω = √50 = 7.071 rad/s.

8.Stability: Routh–Hurwitz & root locus

Answer: si = −18.94

Root locus: as the gain K goes from 0 to ∞, the closed-loop poles, the roots of 1 + K·G(s) = 0, move from the open-loop poles to the open-loop zeros or to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m), which start from the centroid σₐ = (Σ poles − Σ zeros)/(n − m). Branches leave or join the real axis where dK/ds = 0. The angle condition ∠G(s) = 180° fixes the shape, and the magnitude condition K = 1/|G(s)| gives the gain at any point of the locus.

Gain along the real axis

K(s) = −(s² + 2s)/(s + 10)

Stationary points

Differentiate with the quotient rule and set the numerator to zero.

(2s + 2)(s + 10) − (s² + 2s) = s² + 20s + 20 = 0
s = −10 ± √80 = −1.056, −18.94

Break-in point

si = −18.94

Answer

si = −18.94

Between the two break points the locus is a circle centred on the zero, with radius √80 = 8.944.

9.Steady-state error & PID control

Answer: K = 500

Steady-state error: in a unity-feedback loop with open-loop transfer function G(s) the error is E(s) = R(s)/(1 + G(s)), and by the final value theorem ess = lims→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are Kp = lim G(s), Kv = lim s·G(s) and Ka = lim s²·G(s), all for s → 0. A step of size A gives ess = A/(1 + Kp), a ramp A·t gives A/Kv and a parabola A·t²/2 gives A/Ka. A type-0 loop follows a step with a finite error, a type-1 loop follows it exactly and a ramp with a finite error.

Required velocity constant

Kv = A/ess = 5/0.1 = 50

Gain

K = 10·Kv = 500

Answer

K = 500

A large gain makes this loop very lightly damped: ζ = 0.2236. That conflict is why compensators are used.

10.Steady-state error & PID control

Answer: %OS = 28.06 %

P, PI and PID control: the controller C(s) = Kp + Ki/s + Kd·s acts on the error. A proportional gain makes the loop faster and reduces the error of a type-0 plant, but never removes it. The integral term adds a pole at s = 0: the system type rises by one, and a step is followed without error. The derivative term adds damping. To choose gains, write the closed-loop characteristic polynomial and compare it with s² + 2ζωns + ωn². Then %OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd with ωd = ωn√(1 − ζ²), and ts ≈ 4/(ζωn).

Characteristic polynomial

s(s + 6) + 16·4 = s² + 6s + 64

ωₙ and ζ

ωₙ = √64 = 8 rad/s
ζ = 6/(2·8) = 0.375

Overshoot

%OS = 100·e−0.375π/√(1 − 0.375²) = 28.06 %

Answer

%OS = 28.06 %
11.Bode plots & margins

Answer: ∠G(jω) = −91.58 °

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Add the angles

∠G = atan(5/13) − 90° − atan(5/12) = 21.04° − 90° − 22.62°

Result

∠G(j5) = −91.58°

Answer

∠G(jω) = −91.58°
12.Bode plots & margins

Answer: GM = 2.995 dB

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ωc = 1/(RC) or R/L, and fc = ωc/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ωc)²) and φ = −atan(ω/ωc). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ωc: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jωc). Phase crossover ω180: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω180)| dB. The closed loop is stable when both margins are positive.

Phase crossover

(ω/4)·(ω/8) = 1 → ω180 = √(4·8) = 5.657 rad/s

Magnitude at phase crossover

|L(jω180)| = K/(ω·√(ω² + 16)·√(ω² + 64)) = K/(4·8·(4 + 8)) = K/384

Gain margin

GM = 20·log₁₀(384/272) = 2.995 dB

Answer

GM = 2.995 dB

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